Problem: Consider the parametric curve: $\begin{aligned} x&=2t^{-2} \\\\ y&=\ln(t^3) \end{aligned}$ Which integral gives the arc length of the curve over the interval from $t=2$ to $t=7$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\int_{2}^{7} \sqrt{\dfrac{16}{t^{6}}+\dfrac{1}{t^6}}\,dt$ (Choice B) B $\int_{2}^{7} \sqrt{\dfrac{2}{t^2}+\ln(t^3)}\,dt$ (Choice C) C $\int_{2}^{7} \sqrt{\dfrac{16}{t^{6}}+\dfrac{9}{t^2}}\,dt$ (Choice D) D $\int_{2}^{7} \sqrt{-\dfrac{4}{t^{3}}+\dfrac{3}{t}}\,dt$
This is the formula for the arc length of a parametric curve over the interval $[a, b]$ : $\int_a^b\sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\, dt$ [Where does this formula come from?] Let's find $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ : $\begin{aligned} \dfrac{dx}{dt}&=\dfrac{d}{dt}\left[2t^{-2}\right] \\\\ &=-4t^{-3} \\\\ &=-\dfrac{4}{t^3} \\\\\\ \dfrac{dy}{dt}&=\dfrac{d}{dt}\left[\ln(t^3)\right] \\\\ &=\dfrac{3t^2}{t^3} \\\\ &=\dfrac{3}{t} \end{aligned}$ Now we can find the expression for the arc length: $\begin{aligned} &\phantom{=}\int_{2}^{7} \sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\,dt \\\\ &=\int_{2}^{7} \sqrt{\left(-\dfrac{4}{t^3}\right)^2+\left(\dfrac{3}{t}\right)^2}\,dt \\\\ &=\int_{2}^{7} \sqrt{\dfrac{16}{t^{6}}+\dfrac{9}{t^2}}\,dt \end{aligned}$ In conclusion, this integral gives the arc length of the curve over the interval from $t=2$ to $t=7$ : $\int_{2}^{7} \sqrt{\dfrac{16}{t^{6}}+\dfrac{9}{t^2}}\,dt$